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Difference between revisions of "Specimen Cyprus Seniors Provincial/2nd grade/Problem 4"

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== Solution ==
 
== Solution ==
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Since <math>\rho_1, \rho_2 \neq -1</math>, we can multiply both sides by <math>x+1</math>, and <math>\rho_{1}, \rho_{2}</math> will still satisfy the equation:
  
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<math>(x^2-x+1)(x+1)=(0)(x+1)</math>
  
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<math>x^3 + 1 = 0</math>
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<math>x^3 = -1</math>
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Thus, <math>\rho_{1}^3=\rho_{2}^3 = -1</math> as desired <math>\boxed{\mathbb{Q.E.D.}}</math>.
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The third roots of <math>-1 = cis(180^\circ)</math> are <math>cis(60^\circ)</math>, <math>cis(-60^\circ)</math>, and <math>-1</math>.
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Since <math>\rho_1, \rho_2 \neq -1</math>, we have <math>\rho_{1} = cis(60^\circ)</math> and <math>\rho_{2} = cis(-60^\circ)</math>. (Switching the two values will not affect the result).
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Note that <math>\rho_{1}^6=\rho_{2}^6 = 1</math>, and that <math>2006 = 6(334)+2</math>
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So, <math>\rho_{1}^{2006} + \rho_{2}^{2006} = \rho_{1}^{2} + \rho_{2}^{2} = cis(120^\circ) + cis(-120^\circ) = </math> <math>\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) = \boxed{-1}</math>
  
 
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Latest revision as of 23:28, 22 May 2009

Problem

If $\rho_{1}, \rho_{2}$ are the roots of equation $x^2-x+1=0$ then:

a) Prove that $\rho_{1}^3=\rho_{2}^3 = -1$ and

b) Calculate the value of: $\rho_{1}^{2006} + \rho_{2}^{2006}$.


Solution

Since $\rho_1, \rho_2 \neq -1$, we can multiply both sides by $x+1$, and $\rho_{1}, \rho_{2}$ will still satisfy the equation:

$(x^2-x+1)(x+1)=(0)(x+1)$

$x^3 + 1 = 0$

$x^3 = -1$

Thus, $\rho_{1}^3=\rho_{2}^3 = -1$ as desired $\boxed{\mathbb{Q.E.D.}}$.


The third roots of $-1 = cis(180^\circ)$ are $cis(60^\circ)$, $cis(-60^\circ)$, and $-1$.

Since $\rho_1, \rho_2 \neq -1$, we have $\rho_{1} = cis(60^\circ)$ and $\rho_{2} = cis(-60^\circ)$. (Switching the two values will not affect the result).

Note that $\rho_{1}^6=\rho_{2}^6 = 1$, and that $2006 = 6(334)+2$

So, $\rho_{1}^{2006} + \rho_{2}^{2006} = \rho_{1}^{2} + \rho_{2}^{2} = cis(120^\circ) + cis(-120^\circ) =$ $\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) = \boxed{-1}$

See also

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